[momoho]

I can't understand why choice C for part 3 is correct?Anybody can help me?

[mjc123]

As no actual thermodynamic data are given, we can't decide, and you would think the answer would be D. However, since the question says the cell was "designed" in that form, and electrochemical cells are conventionally written with the anode on the left and the cathode on the right, we must assume that the designer knew (or expected) that C2 would have a higher electrode potential than C1, otherwise it would have been written the other way round.

[Big-Daddy]

Well strictly speaking electromotive force in a galvanic cell must be, I think, positive because the EMF measured is a magnitude. So when it says "the electromotive force E of the above cell" it is referring to the EMF measured on the written cell, and EMF measured for any galvanic cell will be positive.

That does not mean that the thermodynamic concept of E with which we use electrode potential to consider Gibbs' free energy of reactions - represented by cell diagrams maybe but not in themselves cells, rather: reactions - cannot be both positive and negative, as with the Gibbs' free energy itself.

So, the bottom-line is: EMF measured, and electromotive force registered on any cell, is always going to be positive for any cell (providing it is not externally interfered with); it's only when looking at electrode potentials for reactions, thermodynamically, that we consider the direction and thus that we call the potential either negative or positive.

So I thought?

[mjc123]

The question defines E as φR - φL, so in principle it could be negative, depending which way round the electrodes are.
EMF has direction as well as magnitude. If you put a galvanometer across the terminals of the cell it will give you a sign as well as a number.

[Big-Daddy]

The question defines E as φR - φL, so in principle it could be negative, depending which way round the electrodes are.
EMF has direction as well as magnitude. If you put a galvanometer across the terminals of the cell it will give you a sign as well as a number.

Yes and I imagine the sign would always be positive unless you apply your own potential externally. I quote Vogel here:

"When a galvanic cell is constructed, a potential difference is measurable between the two electrodes. If the flow of current is negligible, this potential difference is equal to the electromotive force (EMF) of the cell.

EMF = |E₁ - E₂| "

(Vogel's Textbook of Macro and Semimicro Qualitative Inorganic Analysis, 5e, p115)

And that is always the impression I got too - it doesn't make much sense that you switch around in space the sides of a voltmeter and all of a sudden get a different pd...

[mjc123]

Yes and I imagine the sign would always be positive unless you apply your own potential externally

Try it. Attach the black terminal of a voltmeter to the left hand electrode of a galvanic cell and the red terminal to the right hand electrode. Then reverse the connections: black-right, red-left. Do you expect the same sign?

EMF = |E1 - E2|

Vogel is simply using a different definition from the one in our question. E = φR - φL can be positive or negative; which it is tells you which way the (thermodynamically) spontaneous reaction goes.

[Natalia]

I am most likely wrong, but I was thinking, what if hammering makes Cu(1) more dense and conductive and therefore more prone to accepting electrons? I remember hearing somewhere that reduction potential increases (in magnitude) with conductivity, but could someone elaborate on this, if you know anything about it and if you think it relates to the thread?

(if it's true, then it makes sense for E to be greater than zero)

[Arkcon]

I am most likely wrong, but I was thinking, what if hammering makes Cu(1) more dense and conductive and therefore more prone to accepting electrons?

Do you have a source for this phenomena?  If you don't and just made it p, maybe you should ask about it in another thread, and see if an expert can comment.  This thread is in a forum dedicated to standardized tests.  People coming here aren't looking for further complications

I remember hearing somewhere that reduction potential increases (in magnitude) with conductivity, but could someone elaborate on this, if you know anything about it and if you think it relates to the thread?

(if it's true, then it makes sense for E to be greater than zero)

Possible, but I for one don't even understand the concept of hammering two pieces together as changing anything from just one single piece of copper. 

[Natalia]

Do you have a source for this phenomena? 

Admittedly, my only source was intuition - hammering the metal could make atoms stick closer together... or?

Possible, but I for one don't even understand the concept of hammering two pieces together as changing anything from just one single piece of copper. 

Well, assuming that the hammered piece (Cu(1), left in the cell diagram) has a more negative reduction potential than the un-hammered piece (Cu(2), right in the cell diagram), then the electromotive force E = E(Right) - E(Left) would be positive, like the answer C says.

But then again, my only source is intuition. Let me know when you find a satisfactory answer.

[Big-Daddy]

Admittedly, my only source was intuition - hammering the metal could make atoms stick closer together... or?

AFAIK you cannot just (significantly) change the density of a metal by hammering it - you can only deform it that way.

And even if the density is increased, I'm unclear why that should effect the reduction potential in this way?

[burlogata]

Well, there is just one answer- the question is inaccurate and the answer should be E=0; You can't change the chem-properties of Cu piece with hammering it. The Cu(1) = Cu(2) in any chem-properties, so there could not be any potential difference between them.